Question

In: Statistics and Probability

A recent study found that 64 children who watched a commercial for potato chips featuring a...

A recent study found that 64 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 36 grams of potato chips as compared to a mean of 26 grams for 54 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the​ celebrity-endorsed commercial was 21.1 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.9 grams. Complete parts​ (a) through​ (c) below.

a. Assuming that the population variances are equal and alphaequals0.05​, is there evidence that the mean amount of potato chips eaten was significantly higher for the children who watched the​ celebrity-endorsed commercial? Let population 1 be the weights of potato chips eaten by children who watched the​ celebrity-endorsed commercial and let population 2 be the weights of potato chips eaten by children who watched the alternative food snack commercial. What are the correct null and alternative​ hypotheses? a)What is the test​ statistic? (Round to two decimal places)

what is the p value? (round to three decimal places)

b) Assuming that the population variances are​ equal, construct a 95​% confidence interval estimate of the difference mu 1 minus mu 2 between the mean amount of potato chips eaten by the children who watched the​ celebrity-endorsed commercial and children who watched the alternative food snack commercial.determine the 95​% confidence interval using​ technology, rounding to two decimal places.

c) Compare and discuss the results of​ (a) and​ (b). Check to make sure the results of​ (a) and​ (b) can be compared. Recall that confidence intervals are comparable only to the results of a​ two-tail hypothesis test. Since the test done in part​ (a) is a​ one-tail hypothesis​ test, these two results cannot be meaningfully compared.

Solutions

Expert Solution

a.
Given that,
mean(x)=36
standard deviation , s.d1=21.1
number(n1)=64
y(mean)=26
standard deviation, s.d2 =129
number(n2)=54
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.66
since our test is right-tailed
reject Ho, if to > 1.66
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (63*445.21 + 53*16641) / (118- 2 )
s^2 = 7845.0106
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=36-26/sqrt((7845.0106( 1 /64+ 1/54 ))
to=10/16.3663
to=0.611
| to | =0.611
critical value
the value of |t α| with (n1+n2-2) i.e 116 d.f is 1.66
we got |to| = 0.611 & | t α | = 1.66
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: right tail -ha : ( p > 0.611 ) = 0.27119
hence value of p0.05 < 0.27119,here we do not reject Ho
ANSWERS
---------------
i.
null, Ho: u1 = u2
alternate, H1: u1 > u2
ii.
test statistic: 0.611
critical value: 1.66
decision: do not reject Ho
iii.
p-value: 0.27119
we do not have enough evidence to support the claim that he mean amount of potato chips eaten was significantly higher for the children who watched the​ celebrity-endorsed commercial
b.
TRADITIONAL METHOD
given that,
mean(x)=36
standard deviation , s.d1=21.1
number(n1)=64
y(mean)=26
standard deviation, s.d2 =129
number(n2)=54
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (63*445.21 + 53*16641) / (118- 2 )
s^2 = 7845.011
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 7845.011 * (1/64+1/54) )
=16.366
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 116 d.f is 1.981
margin of error = 1.981 * 16.366
= 32.422
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (36-26) ± 32.422 ]
= [-22.422 , 42.422]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=36
standard deviation , s.d1=21.1
sample size, n1=64
y(mean)=26
standard deviation, s.d2 =129
sample size,n2 =54
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 36-26) ± t a/2 * sqrt( 7845.011 * (1/64+1/54) ]
= [ (10) ± 32.422 ]
= [-22.422 , 42.422]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-22.422 , 42.422]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
c.
ANSWERS:
a.
we do not have enough evidence to support the claim that he mean amount of potato chips eaten was significantly higher for the children who watched the​ celebrity-endorsed commercial
b.
95% sure that the interval [-22.422 , 42.422]
confidence intervals are comparable only to the results of a two-tail hypothesis test.
part (a) is one tailed test.


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